Redshift Between Emission and Reception

A redshift compares two local frequency measurements linked by one propagating light signal. The emitter and receiver contribute their four-velocities at different events, while the null wave vector carries the phase information between them. The exact ratio is coordinate independent, but splitting it into “Doppler” and “gravitational” pieces requires an additional reference frame or spacetime symmetry.

Required background. Observers, Rest Spaces, and Measured Components provides observer projectors and measured decompositions; Pseudo-Riemannian Metrics and Signature provides Lorentzian norms and causal classification; Frames, Coframes, and Anholonomy provides dual frames and local Lorentz freedom.

Helpful background. Events, Intervals, and Minkowski Geometry supplies the flat-spacetime limit; Tensor Fields and Natural Operations clarifies tensors versus their components; Killing Fields and Isometry Groups supplies the stationary conserved quantity; The Wave Equation and Characteristics connects the geometric-optics phase to its transported wave vector.

The endpoint frequency ratio

Let a geometric-optics light ray connect emission event $e$ to reception event $r$. Its future-directed null wave vector is $k^\mu$, with the same phase normalization transported along the ray. If the emitter and receiver have four-velocities $u_e^\mu$ and $u_r^\mu$, each measures

$$\omega_e=-(k_\mu u_e^\mu)_e, \qquad \omega_r=-(k_\mu u_r^\mu)_r.$$

Define the redshift by $1+z=\lambda_r/\lambda_e=\omega_e/\omega_r$. Therefore

$$1+z =\frac{(-k\cdot u)_e}{(-k\cdot u)_r}.$$

The signs are essential with $u^2=-1$: both contractions are positive for a future-directed wave and future-directed observer. A positive $z$ means the received frequency is smaller. An overall constant rescaling of $k^\mu$ cancels between endpoints, although a physical wave train fixes its normalization through the phase.

The ratio also has a direct clock interpretation. Follow two neighboring wave crests from the emitter to the receiver along the same smoothly varying ray branch. The number of phase cycles is unchanged during propagation, so the magnitudes of the phase increments satisfy

$$\omega_e\,d\tau_e=\omega_r\,d\tau_r, \qquad 1+z=\frac{d\tau_r}{d\tau_e}.$$

A redshift therefore means a longer received interval between successive crests. This is a differential comparison of nearby pulses; a long evolving pulse train need not have one constant redshift.

The formula assumes geometric optics and a specified null path. If lensing produces several rays between the same worldlines, each image can have a different $k^\mu$ and hence a different redshift and arrival time. If there is no connecting ray, no such comparison exists.

Motion and propagation factors

There is no invariant, observer-free relative velocity between separated events. To organize the result, choose a smooth reference observer field $n^\mu$ at the two endpoints. Multiplying by unity gives the exact factorization

$$1+z =\frac{(-k\cdot u_e)_e}{(-k\cdot n)_e} \frac{(-k\cdot n)_e}{(-k\cdot n)_r} \frac{(-k\cdot n)_r}{(-k\cdot u_r)_r}.$$

The first and third factors are local Doppler comparisons between the physical endpoint observers and $n^\mu$. The middle factor is the frequency change measured by the chosen reference congruence along propagation. A different $n^\mu$ redistributes the factors while leaving their product fixed. Thus the total redshift is observable, but a generic motion-versus-geometry split is not unique.

In flat spacetime, take an inertial reference field and a photon moving in the $+x$ direction. A receiver receding at speed $v$ measures $\omega_r=\gamma(1-v)\omega_e$, so

$$1+z=\sqrt{\frac{1+v}{1-v}}.$$

This special-relativistic limit checks the endpoint signs: recession gives $z>0$, approach gives $z<0$.

Stationary spacetime and Killing energy

Suppose the relevant region admits a timelike Killing field $\xi^\mu$. Along an affinely parametrized null geodesic,

$$\mathcal E=-\xi_\mu k^\mu$$

is constant, because $k^\nu\nabla_\nu(\xi_\mu k^\mu)=k^\mu k^\nu\nabla_{(\mu}\xi_{\nu)}=0$. Observers following the Killing flow have

$$u^\mu=\frac{\xi^\mu}{N}, \qquad N=\sqrt{-\xi^\mu\xi_\mu},$$

so they measure $\omega=\mathcal E/N$. For a stationary emitter and receiver connected by the ray,

$$1+z=\frac{N_r}{N_e}.$$

The normalization of $\xi^\mu$ cancels from this ratio. The formula applies only where the chosen Killing field is timelike and the observers actually follow its integral curves. It is not a general formula for dynamical spacetimes, rotating observers not aligned with $\xi^\mu$, or arbitrary moving endpoints.

In a weak, static field with $g_{00}=-(1+2\Phi)+O(\Phi^2)$, $N=1+\Phi+O(\Phi^2)$ and

$$z=\Phi_r-\Phi_e+O(\Phi^2).$$

A signal climbing to larger Newtonian potential is redshifted. The equivalence-principle derivation and curved-clock interpretation are given in Reall 2013, Ch. 1, pp. 6–10, PDF and Carroll 1997, Ch. 4, pp. 101–104.

Interpretation and limits

Redshift compares clock rates only after a light-transfer protocol and source frequency standard are specified. It is not an intrinsic property located “inside the photon,” nor does it define a global distance. Absorption, dispersion, plasma propagation, finite wave packets, source evolution, detector response, and failure of geometric optics can alter what a real spectrometer infers; those effects lie beyond the vacuum null-ray model used here.

Even in vacuum, a conserved coordinate component such as $k_t$ has physical meaning only when tied to a symmetry and its normalization. The endpoint contractions remain the general statement. They cleanly separate what is measured locally from what must be propagated through the spacetime between measurements.

Exercises

In a static region, two observers follow the same timelike Killing field at lapse values $N_e$ and $N_r=2N_e$. A monochromatic signal is emitted by the first and received by the second. Find the frequency ratio and redshift.

Solution

The conserved Killing energy gives $\omega_e=\mathcal E/N_e$ and $\omega_r=\mathcal E/N_r=\omega_e/2$. Hence

$$1+z=\frac{\omega_e}{\omega_r}=2, \qquad z=1.$$

The receiver measures half the emitted frequency. The result assumes both observers are stationary with respect to the same Killing flow and that the connecting null ray stays within the stationary region.

References

• Carroll, Sean M. “Lecture Notes on General Relativity”, arXiv:gr-qc/9712019v1, 1997, Ch. 4, pp. 101–104.
• Reall, Harvey S. General Relativity, PDF, Part III lecture notes, University of Cambridge, 2013, Ch. 1, pp. 6–10.